Urgent: Maths CIE Stats

[cs]

June 2010 Q1,

how do you know if that question is asking for binomial or P and C..?

A bottle of sweets contains 13 red sweets, 13 blue sweets, 13 green sweets and 13 yellow sweets. 7 sweets are selected at random . Find the probability that exactly 3 of them are red.

My first instinct is to do X~B (7, 13/52)
and then P(x=3) then continue,

but the answer is found using 13C3 X 39C4  over 52C7

anyone can help me?

[Deadly_king]

June 2010 Q1,

how do you know if that question is asking for binomial or P and C..?

A bottle of sweets contains 13 red sweets, 13 blue sweets, 13 green sweets and 13 yellow sweets. 7 sweets are selected at random . Find the probability that exactly 3 of them are red.

My first instinct is to do X~B (7, 13/52)
and then P(x=3) then continue,

but the answer is found using 13C3 X 39C4  over 52C7

anyone can help me?

Nope.......you cannot use binomial here. This is because you don't even now what are the colours of the selected sweets. So you cannot take if fro granted.

Moreover the word exactly gives you a hint that you should be using combinations since the selected sweets must be 3 red and the rest a ny other colour.

So firstly, the probability of 3 red ---> 13C3

Since you have been told that there is exactly 3 red and not more, this implies that the rest(4) should be of other colours out of only 39 sweets ---> 39C4

Since you are finding probability, you should divide the above product by the probability of taking out 7 sweets = 52C7

Hence the solution is (13C3 x 39C4)/52C7

[cs]

Thank you Deadly King, you are always here to solve my doubt.. =)

[mousa]

I would like to add more, the probability of success in " choosing a red sweet " is NOT constant, it keeps on changing everytime you make a selection since the sweets are not replaced. Furthermore, the trials aren't independent. Both of those conditions are essential for binomial distribution.

hope that helped, by the way, your doing your exam in this nov session??

[Deadly_king]

Thank you Deadly King, you are always here to solve my doubt.. =)

Anytime dear

By the way when are you having your stats exams 3 Nov

Thanks for your info mate
Very much appreciated

[cs]

Thanks both of you, yes i am sitting for the exam on 3 Nov, good luck!

[Deadly_king]

Thanks both of you, yes i am sitting for the exam on 3 Nov, good luck!

Same day for me as well

Though I'll be sitting for 62

[mousa]

Same day for me as well

Though I'll be sitting for 62

hahaha meeee 2 !!! i will be sitting 62 , where are you  living??

[Deadly_king]

hahaha meeee 2 !!! i will be sitting 62 , where are you  living??

Hehe......that's nice

Am from Mauritius..............and you

[mousa]

Hehe......that's nice

Am from Mauritius..............and you

Wowww nice, muritaniaaaa loool, thats tooo far from here I am Jordanian living in the UAE, so you are Muratious ya3ny??

[Deadly_king]

Wowww nice, muritaniaaaa loool, thats tooo far from here I am Jordanian living in the UAE, so you are Muratious ya3ny??

Hehe........Not Mauritania but Mauritius, tiny island in the Indian Ocean

[$!$RatJumper$!$]

I would like to add more, the probability of success in " choosing a red sweet " is NOT constant, it keeps on changing everytime you make a selection since the sweets are not replaced. Furthermore, the trials aren't independent. Both of those conditions are essential for binomial distribution.

hope that helped, by the way, your doing your exam in this nov session??

When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.

[Deadly_king]

When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.

The trials are NOT independent ----> each trial depends on the previous trial since the previous trial decreased the total number of sweets in the bag, so the probability will depend on how many sweets have been removed and have they been replaced or not.

[cs]

Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2

why 6! X 7C3?

[Deadly_king]

Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2

why 6! X 7C3?

I can't really explain the workings from the marking scheme but I did it another way which was approved by my teacher.

First we find out the number of arrangements in which all the 9 persons can be arranged. This will be (9! +(7! x 3!))
9! => There are nine persons which can be arranged in any way.
(7! x 3!) => There are 6 men and 3 women. We'll be taking all the women to be standing beside each other. Hence they are taken as one possiblity only which makes 7!. The 3 women can then be arranged in any way 3!.

Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
8! = > One woman is taken as one possiblity and the other two as another possiblity. In all there will be 8 possiblities, after adding the six men.


Number of arrangements : (9! +(7! x 3!)) - 3(8! x 2!) = 151200

Hope you understand